Question

[Nguyễn Thùy Trang]

Bài 11:
Tính \(\frac{\left(1986^2-1992\right)\left(1986^2+3972-3\right).1987}{1983.1985.1988.1989}\)

Bài 12: Đặt \(a+b+c=2p\). CMR:

\(\frac{1}{p-a}+\frac{1}{p-b}+\frac{1}{p-c}-\frac{1}{p}=\frac{abc}{p\left(p-a\right)\left(p-b\right)\left(p-c\right)}\)

Answer 1

[Natsu Dragneel]

Bài 11 : ta có :

\(\frac{\left(1986^2-1992\right)\left(1986^2+3972-3\right)1987}{1983.1985.1988.1989}\)

\(=\frac{\left(1986^2-3.1986+2.1986-6\right)\left(1986^2+2.1986+1-4\right)1987}{1983.1985.1988.1989}\)

\(=\frac{\left(1986-3\right)\left(1986+2\right)\left[\left(1986+1\right)^2-2^2\right]1987}{1983.1985.1988.1989}\)

\(=\frac{1983.1988\left(1987-2\right)\left(1987+2\right)1987}{1983.1988.1985.1989}\)

\(=\frac{1983.1985.1988.1989.1987}{1983.1985.1988.1989}=1987\)